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4 years ago

9

If you install special sound-reflecting windows that reduce the sound intensity level by 33.0 dBdB , by what factor have you red

uced the sound intensity

1 answer:

Sever21 [200]4 years ago

3 0

Answer:

The sound intensity is reduced by a factor of 1995.26

Explanation:

When comparing two sound intensities, the intensity level is measured in the unit of decibel or dB. The intensity of the threshold of hearing for a human being is 10^−12 W/m^2 . When the intensity level is zero, it means that the sound intensity is the same as the threshold of hearing.

The reduced sound intensity level is given as 33 dB, so

10 log (I/Io) = - 33

I : intensity of the sound

Io ( = 10^−12 W/m^2): threshold of hearing

So, the intensity ratio is

I/Io = 10^-3.3

= 5.01 x 10^-4

1/ 5.01 x 10^-4 = 1995.26

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